Q>K

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Aishwarya Kosgi 1F
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Q>K

Postby Aishwarya Kosgi 1F » Fri Jan 08, 2021 10:12 am

Hi! I had a question about when the reaction quotient is larger than the equilibrium constant. I was wondering how this was possible because I figured once the reaction reached equilibrium the concentrations would remain the same. I wasn't sure how it would be possible to go beyond the equilibrium value and have a Q value greater than K. If someone could explain how it is possible to get a Q value greater than K, that would be great! Thank you!

Steph Du 1H
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Re: Q>K

Postby Steph Du 1H » Fri Jan 08, 2021 10:27 am

A possibility could be if you had a reaction and then you added a bunch of products? So that would lead to a greater Q value and then the reaction would proceed in a reverse reaction.

AnjikaFriedman-Jha2D
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Re: Q>K

Postby AnjikaFriedman-Jha2D » Fri Jan 08, 2021 10:50 am

I think that there are many instances where the reaction does not reach equilibrium rather it just shifts back and forth between favoring the forward or reverse reaction

Neha Gupta 2A
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Re: Q>K

Postby Neha Gupta 2A » Fri Jan 08, 2021 11:22 am

Another way that Q>K is if you had a system at equilibrium and then removed a bunch of reactants so that [products] > [reactants]

BKoh_2E
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Re: Q>K

Postby BKoh_2E » Fri Jan 08, 2021 11:44 am

You are correct that a reaction at equilibrium will not naturally make more products and have a Q value greater than the equilibrium constant. Examples of when Q will be higher are due to external factors (reactants have been removed) or the reaction has not yet reached equilibrium .

Eliana Carney 3E
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Re: Q>K

Postby Eliana Carney 3E » Sun Jan 10, 2021 9:18 am

Hey Aishwarya!

You are correct in saying that once a reaction reaches equilibrium the concentrations of products and reactants would stay the same. Essentially, when Q>K, some outside occurrence has occurred that has shifted the reaction out of equilibrium. For example, a change in temperature, the removal of reactants, or the addition of products could potentially cause Q>K. Hope this helps!

Shalyn Kelly 3H
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Re: Q>K

Postby Shalyn Kelly 3H » Sun Jan 10, 2021 9:47 am

Just to clarify, if you remove enough reactants the system would likely be Q>K (where [products]>[reactants]) and if you add enough reactants the system could tip to be Q<K ([products]<[reactants])?

Chinyere Okeke 2J
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Re: Q>K

Postby Chinyere Okeke 2J » Sun Jan 10, 2021 10:42 am

Shalyn Kelly 3H wrote:Just to clarify, if you remove enough reactants the system would likely be Q>K (where [products]>[reactants]) and if you add enough reactants the system could tip to be Q<K ([products]<[reactants])?


Hi! I believe that you are correct as when you remove reactant, the reaction will shift so that it can produce more reactant, so the reverse reaction will be favored in this case. And it would be Q<K ([products]<[reactants]) since when this occurs there would be more products than reactants.

Chinyere Okeke 2J
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Re: Q>K

Postby Chinyere Okeke 2J » Sun Jan 10, 2021 10:45 am

I think you are correct as when a reaction is at equilibrium there won't be a net increase in product concentration so a Q value greater than the equilibrium constant, K, therefore there would have to be external factors or the reaction has not yet reached equilibrium.

Isabel_Eslabon_2G
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Re: Q>K

Postby Isabel_Eslabon_2G » Sun Jan 10, 2021 10:49 am

I think one example of when Q > K is when the reaction is not quite at equilibrium yet so there are still products that are forming reactants.

Because K & Q = [Products]/[Reactants], if Q > K, then there are more products that are still forming those reactants. This can also occur if more products are added (numerator becomes larger) or some reactants are removed (denominator becomes smaller) once the reaction reached equilibrium.

Maddie Turk Disc 2J
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Re: Q>K

Postby Maddie Turk Disc 2J » Sun Jan 10, 2021 7:02 pm

So this is just when the reaction is no longer at equilibrium anymore?


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